3
$\begingroup$

This is the first time I ask a question here, so sorry if I make any mistake in the way I ask it. I'm studing Fefferman's article Pointwise Convergence of Fourier Series, and I have two questions:

1) In the first page, it is defined a function $n(\cdot)$. How can I prove that it is measurable?

2) In the third page, I understand the proof of $||T_p||_2\leq CA_0(p)^{1/2}$. The author says that it is easy to prove the opposite inequality. How can I manage to do that?

3) The author considers intervals $I$ or $\omega$ and denotes by $I^*$ and $\omega^*$ the double of $I$ and $\omega$ respectively. What is exactly the double of an interval?

Thank you for your answers.

$\endgroup$
2
  • 1
    $\begingroup$ I have had similar concerns when studying the pointwise convergence problem for the Schrödinger group. I found the answer to the analogue of your question 1) in this MathSE post, and then I got useful information in this MO post. Hope this helps $\endgroup$ – Giuseppe Negro Jan 27 '16 at 16:37
  • $\begingroup$ 7 of the 8 sections are about $L^2$ $\endgroup$ – john mangual Sep 1 '17 at 15:26
1
$\begingroup$

For your first question, as Fefferman notes, it suffices to assume that $n(x)$ is bounded assuming the constants in the final result (inequality (1)) do not depend on this bound. This is equivalent to saying that it suffices to prove inequality (1) for trigonometric series of length $M$, as long as the constant in (1) doesn't depend on $M$. Now by the uncertainty principle (or, say, Bernstein's inequality) a trigonometric series of length $M$ is nearly constant on intervals of length $1/M$. Using these observations, it isn't too hard to see that it suffices to consider integer-valued (simple) functions $n(x) : \mathbb{T} \rightarrow [1,2,\ldots,M]$ that are constant intervals of $1/M$ which are clearly measurable.

[For 2) can you post the definitions of $A_0(p)$ and $T_p$?]

$\endgroup$
5
  • $\begingroup$ Dear Mark Lewko, thank you for your answer. I give the definition of $T_p$. On $[0,2\pi]$ and considering the Lebesgue measure $dx/2\pi$, the author defines a pair $[\omega,I]$ as the dyadic intervals $\omega\subseteq R$ and $I\subseteq [0,2\pi]$ such that $|\omega|=1/|I|$. Let $p=[\omega,I]$ be a pair. Then $E(p)$ is defined as $\{x\in I:n(x)\in\omega\}$. Define $T_pf(x)=\int_{-\pi}^{\pi}e^{in(x)y}\psi_k(y)f(x-y)dy \chi_{E(p)}(x)$. $\endgroup$ – jxm Jan 27 '16 at 17:00
  • $\begingroup$ $\{\psi_k\}_k$ is a sequence of functions such that $1/y=\sum\psi_k(y)$ for $|y|<\pi$, $\psi_k(y)=2^k\psi(2^ky)$ and $\psi$ is an odd $C^{\infty}$ function supported in $[-2\pi,2\pi]$. The definition for $A_0(p)$ is $|E(p)|/|I|$. The author proves $||T_p||_2\leq CA_0(p)^{1/2}$ (I forgot the $1/2$ in the question). $\endgroup$ – jxm Jan 27 '16 at 17:00
  • 1
    $\begingroup$ In the case that $\omega$ is centred at the origin, one can simply test $T_p$ with the function $f = 1_I$, the point being that the phase $e^{in(x) y}$ is close to 1. The general case is a frequency modulation of this general case, so one needs to multiply $f$ by an appropriate phase. Also, the double $I^*$ of an interval $I$ usually refers to the interval with the same centre as $I$, but twice the length. $\endgroup$ – Terry Tao Jan 28 '16 at 19:32
  • $\begingroup$ @TerryTao Thank you for your answer. However, I don't understand how I can get rid of the $\psi_k$ to lower-bound $|T_p (f)| $. Could you please provide more details? I graduated last year and I'm not very used to reading this type of articles $\endgroup$ – jxm Feb 10 '16 at 18:27
  • $\begingroup$ @jxm could we just use the property of the Hilbert-Schmidt integral operator instead? $\endgroup$ – Thomas Yang Jun 10 '17 at 14:11
1
$\begingroup$

I have a detailed answer for 1) (a professor of mine helped me with this). Take $M\in\mathbb{N}$ such that $n(x)\leq M$ for all $x\in [0,2\pi]$. Consider the sets $$E_1=\{x\in [0,2\pi]:\sup_{N\leq M}|S_Nf(x)|=|S_1f(x)|\},$$ $$E_2=\{x\in [0,2\pi]\backslash E_1:\sup_{N\leq M}|S_Nf(x)|=|S_2f(x)|\},$$ $$ \cdots $$ $$E_M=\{x\in [0,2\pi]\backslash E_{M-1}:\sup_{N\leq M}|S_Nf(x)|=|S_Mf(x)|\}.$$ These sets are measurable, as a consequence of the following result: ''if $A$ is measurable in $\mathbb{R}$ and $F,G:A\rightarrow\mathbb{R}$ are measurable, then $\{x\in A:F(x)=G(x)\}$ is measurable''.

Now notice that $$n=\sum_{j=1}^Mj\,\chi_{E_{j}},$$ which is measurable.

$\endgroup$
1
$\begingroup$

1) This is not necessary for this particular question, but in my experience the Arsenin-Kunugui measurable selection theorem (see Kechris's book "Classical Descriptive Set Theory") is the "one size fits all" solution to measurable selection questions.

3) Double of an interval = interval with the same center and twice the length.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.